Distance Formula and Section Formula

In coordinate geometry, we represent points on a plane using ordered pairs of numbers (x, y). These coordinates allow us to calculate various properties of geometric shapes, such as distances between points, areas of triangles, and the location of points that divide a line segment in a specific ratio. The distance formula and the section formula are fundamental tools for these calculations.

Distance Formula

The distance formula is derived from the Pythagorean theorem. It helps us find the length of a line segment connecting two points in a Cartesian coordinate system.

Derivation of the Distance Formula

Consider two points, P(x₁, y₁) and Q(x₂, y₂), in a Cartesian plane. We want to find the distance between P and Q, denoted as PQ. We can form a right-angled triangle by drawing a horizontal line from P and a vertical line from Q (or vice-versa) that intersect at a point, say R. The coordinates of R will be (x₂, y₁). The length of the horizontal side PR is the absolute difference in the x-coordinates, which is |x₂ - x₁|. The length of the vertical side QR is the absolute difference in the y-coordinates, which is |y₂ - y₁|. According to the Pythagorean theorem in the right-angled triangle PQR, we have: PQ² = PR² + QR² PQ² = (x₂ - x₁)² + (y₂ - y₁)² Taking the square root of both sides, we get the distance formula: PQ = √((x₂ - x₁)² + (y₂ - y₁)²)

The Distance Formula

The distance between two points (x₁, y₁) and (x₂, y₂) is given by: d = √((x₂ - x₁)² + (y₂ - y₁)²)

Example 1:

Find the distance between the points A(2, 3) and B(5, 7). Here, x₁ = 2, y₁ = 3, x₂ = 5, and y₂ = 7. Using the distance formula: AB = √((5 - 2)² + (7 - 3)²) AB = √(3² + 4²) AB = √(9 + 16) AB = √(25) AB = 5 units.

Example 2:

Calculate the distance between the origin (0, 0) and the point P(-4, 3). Here, x₁ = 0, y₁ = 0, x₂ = -4, and y₂ = 3. Distance OP = √((-4 - 0)² + (3 - 0)²) OP = √((-4)² + 3²) OP = √(16 + 9) OP = √(25) OP = 5 units. This is also the formula for finding the magnitude of a vector from the origin to the point (-4, 3).

Special Cases of the Distance Formula:

• Distance from Origin: The distance of a point P(x, y) from the origin O(0, 0) is √(x² + y²).
• Distance between two points on the x-axis: If two points lie on the x-axis, their coordinates are (x₁, 0) and (x₂, 0). The distance between them is |x₂ - x₁|.
• Distance between two points on the y-axis: If two points lie on the y-axis, their coordinates are (0, y₁) and (0, y₂). The distance between them is |y₂ - y₁|.

Section Formula

The section formula helps us find the coordinates of a point that divides a line segment joining two given points internally or externally in a specified ratio.

Internal Division:

Let P(x, y) be a point that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n. This means that the ratio of the distance AP to PB is m/n. The coordinates of point P are given by the section formula: x = (mx₂ + nx₁) / (m + n) y = (my₂ + ny₁) / (m + n) So, P(x, y) = ( (mx₂ + nx₁) / (m + n), (my₂ + ny₁) / (m + n) )

Derivation of the Section Formula (Internal Division):

Consider points A(x₁, y₁) and B(x₂, y₂). Let P(x, y) divide AB internally in the ratio m:n. Draw perpendiculars from A, P, and B to the x-axis, meeting at points X₁, X, and X₂ respectively. Draw a line through P parallel to the x-axis, meeting AX₁ and BX₂ at points C and D respectively. We have AP/PB = m/n. Consider the triangles APC and PDB. Angle PAC = Angle PBD (alternate interior angles, since AC is parallel to BD and AB is a transversal). Angle ACP = Angle BDP = 90 degrees. Therefore, triangle APC is similar to triangle PDB by AA similarity. The ratio of corresponding sides is equal: AP/PB = AC/PD = PC/BD We know AP/PB = m/n. From the diagram, AC = XX₁ = x - x₁ and PD = X₂X = x₂ - x. Also, PC = CX₁ = X₂X = y - y₁ and BD = DX₂ = y₂ - y. So, m/n = (x - x₁)/(x₂ - x) => m(x₂ - x) = n(x - x₁) => mx₂ - mx = nx - nx₁ => mx₂ + nx₁ = (m+n)x => x = (mx₂ + nx₁) / (m + n) And, m/n = (y - y₁)/(y₂ - y) => m(y₂ - y) = n(y - y₁) => my₂ - my = ny - ny₁ => my₂ + ny₁ = (m+n)y => y = (my₂ + ny₁) / (m + n)

Example 3:

Find the coordinates of the point that divides the line segment joining A(-1, 7) and B(4, -3) internally in the ratio 2:3. Here, x₁ = -1, y₁ = 7, x₂ = 4, y₂ = -3, m = 2, and n = 3. Using the section formula for internal division: x = (2 * 4 + 3 * -1) / (2 + 3) = (8 - 3) / 5 = 5 / 5 = 1 y = (2 * -3 + 3 * 7) / (2 + 3) = (-6 + 21) / 5 = 15 / 5 = 3 The coordinates of the point are (1, 3).

Mid-point Formula:

The mid-point formula is a special case of the section formula where the ratio m:n is 1:1. This means the point divides the line segment into two equal halves. If P(x, y) is the mid-point of the line segment joining A(x₁, y₁) and B(x₂, y₂), then: x = (1 * x₂ + 1 * x₁) / (1 + 1) = (x₁ + x₂) / 2 y = (1 * y₂ + 1 * y₁) / (1 + 1) = (y₁ + y₂) / 2 So, the mid-point P is ((x₁ + x₂) / 2, (y₁ + y₂) / 2).

Example 4:

Find the mid-point of the line segment joining P(3, 8) and Q(7, 2). Using the mid-point formula: Mid-point x = (3 + 7) / 2 = 10 / 2 = 5 Mid-point y = (8 + 2) / 2 = 10 / 2 = 5 The mid-point is (5, 5).

External Division:

Let P(x, y) be a point that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) externally in the ratio m : n. This means that the ratio of the distance AP to PB is m/n, and P lies outside the segment AB. The coordinates of point P are given by the section formula for external division: x = (mx₂ - nx₁) / (m - n) y = (my₂ - ny₁) / (m - n) Note: This formula is valid only if m ≠ n. If m = n, the point P would be at infinity, implying the line is parallel to AB.

Derivation of the Section Formula (External Division):

Consider points A(x₁, y₁) and B(x₂, y₂). Let P(x, y) divide AB externally in the ratio m:n. This means AP/PB = m/n. P is on the line AB but outside the segment. We can treat this as internal division of a line segment where one of the points is effectively "negative". Imagine point A is at (x₁, y₁) and point P is at (x, y). The ratio AP/PB = m/n. We can rewrite this relationship using directed distances. If P divides AB externally in the ratio m:n, then the ratio of distances from P to A and P to B is m:n. Let's consider vectors. Vector PA = A - P, Vector PB = B - P. The ratio AP/PB = m/n implies that P is located such that the vector AP is m times some unit vector and PB is n times the same unit vector, and they are in opposite directions relative to the segment AB. A simpler way to derive it is to consider the ratio m:(-n) for internal division. If P divides AB externally in ratio m:n, then P divides the line joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m:(-n). Using the internal division formula with m' = m and n' = -n: x = (m * x₂ + (-n) * x₁) / (m + (-n)) = (mx₂ - nx₁) / (m - n) y = (m * y₂ + (-n) * y₁) / (m + (-n)) = (my₂ - ny₁) / (m - n)

Example 5:

Find the coordinates of the point that divides the line segment joining A(1, -2) and B(3, 4) externally in the ratio 3:1. Here, x₁ = 1, y₁ = -2, x₂ = 3, y₂ = 4, m = 3, and n = 1. Using the section formula for external division: x = (3 * 3 - 1 * 1) / (3 - 1) = (9 - 1) / 2 = 8 / 2 = 4 y = (3 * 4 - 1 * -2) / (3 - 1) = (12 - (-2)) / 2 = (12 + 2) / 2 = 14 / 2 = 7 The coordinates of the point are (4, 7).

Applications of Distance and Section Formulas

These formulas are crucial in various geometric problems:

• Verifying geometric shapes: We can use the distance formula to check if three points are collinear (if the sum of the distances between two pairs of points equals the distance between the third pair). We can also use it to find the lengths of sides of triangles, squares, rectangles, etc., and their diagonals to classify them.
• Finding centroids of triangles: The centroid is the intersection of the medians, and its coordinates can be found using a variation of the section formula.
• Trigonometric problems: Calculating distances can be a step in solving problems involving angles and lengths in coordinate geometry.
• Physics: Calculating displacement, distance between objects, and positions in a 2D or 3D space.

Collinearity of Three Points:

Three points A, B, and C are collinear if the sum of the lengths of any two line segments among AB, BC, and AC is equal to the length of the remaining line segment. For example, if AB + BC = AC, then A, B, and C are collinear, with B lying between A and C.

Example 6:

Check if the points P(1, 1), Q(3, 3), and R(5, 5) are collinear. First, calculate the distances between each pair of points using the distance formula. PQ = √((3-1)² + (3-1)²) = √(2² + 2²) = √(4 + 4) = √(8) = 2√2 QR = √((5-3)² + (5-3)²) = √(2² + 2²) = √(4 + 4) = √(8) = 2√2 PR = √((5-1)² + (5-1)²) = √(4² + 4²) = √(16 + 16) = √(32) = 4√2 Now, check if the sum of two distances equals the third: PQ + QR = 2√2 + 2√2 = 4√2 This is equal to PR. Therefore, the points P, Q, and R are collinear.

Centroid of a Triangle:

If the vertices of a triangle are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃), the coordinates of its centroid G are given by: G = ( (x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3 ) This formula is derived by considering that the centroid divides each median in a 2:1 ratio.

Example 7:

Find the centroid of a triangle with vertices P(2, 5), Q(4, 1), and R(6, 9). Using the centroid formula: Centroid x = (2 + 4 + 6) / 3 = 12 / 3 = 4 Centroid y = (5 + 1 + 9) / 3 = 15 / 3 = 5 The centroid of the triangle is (4, 5).